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I was doing an exercise and I found that for example:

| x - 2/3|= 2

x is 8/3 or -4/2 because x less 2/3 needs to be 2 or -2 because the distance between 2 or -2 to 0 is the same (2).

But when applying this to an inequation I don't understand the meaning. For example:

|x| < 3

How can I solve that?

I would appreciate if someone could help me understanding that.

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$$\left| x-\frac { 2 }{ 3 } \right| =2\quad \Rightarrow \begin{cases} x-\frac { 2 }{ 3 } =2 \\ x-\frac { 2 }{ 3 } =-2 \end{cases}\Rightarrow \begin{cases} x=\frac { 8 }{ 3 } \\ x=-\frac { 4 }{ 3 } \end{cases}\\ \left| x \right| <3\Rightarrow \quad -3<x<3$$

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  • $\begingroup$ Could you please explain how you get that result? (-3 < x < 3) $\endgroup$ – Francisca Sousa Oct 23 '16 at 12:09
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    $\begingroup$ It might help if you understand that with $x\in \mathbb{R}$ , $|x|$ is the distance to $0$. In your example $|x|<3$ means the distance from $x$ to $0$ is less than $3$, thus $x$ can take any values between $-3$ and $3$, ie : $-3<x<3$ $\endgroup$ – Furrane Oct 23 '16 at 12:14
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You have to solve $x=0$ (the expression inside the absolute value equal to zero). Now, have in mind that $$|x|=\begin{cases}x & x\ge 0 \\ -x & x<0\end{cases}$$ Then you study the inequalities $x<3$ and $-x<3.$

Consider another example: $|2x+3|<5.$ Solve $2x+3=0$ and get $x=-2/3.$ If $x\ge-2/3$ solve $2x+3<5$ and if $x<-2/3$ solve $-2x-3<5.$

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